Question

# The vector from the point $$-\hat{i} + 2\hat {j} + 6\hat{k}$$ to the straight line through the point $$(2, 3, -4)$$ and parallel to the vector $$6\hat{i} + 3\hat {j} + 4\hat { k}$$, is

A
29761i+11861j53461k
B
29761i11861j53461k
C
29761i+11861j+53461k
D
none of these

Solution

## The correct option is A $$\dfrac{297}{61}i+\dfrac{118}{61}j-\dfrac{534}{61}k$$Equation of the straight line will be $$\dfrac{x-2}{6}=\dfrac{y-3}{3}=\dfrac{z+4}{4}=t$$Therefore$$x=6t+2$$$$y=3t+3$$$$z=4t-4$$Therefore, let any point P on the line be $$P=(6t+2,3t+3,4t-4)$$And $$A=(-1,2,6)$$Therefore $$\vec{AP}=(6t+3)i+(3t+1)j+(4t-10)k$$Now for perpendicular distance$$\vec{AP}.(6i+3j+4k)=0$$Or $$((6t+3)i+(3t+1)j+(4t-10)k).(6i+3j+4k)=0$$$$6(6t+3)+3(3t+1)+4(4t-10)=0$$$$36t+18+9t+3+16t-40=0$$$$61t-19=0$$$$t=\dfrac{19}{61}$$Hence $$\vec{AP}=\dfrac{297}{61}i+\dfrac{118}{61}j-\dfrac{534}{61}k$$Hence $$d=|\vec{AP}|$$Mathematics

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